Generalized Hooke’s law and allowable stress

A steel bracket rarely feels stress in just one direction. Pull on a flat plate along xx and it also pushes or pulls along yy and zz because of the material's own Poisson contraction — the simple σ=Eε\sigma = E\varepsilon you learned for a bar in tension cannot capture that coupling. The generalized Hooke's law fixes this: it superposes the direct strain from each stress component with the sideways Poisson contraction from the other two, giving the full strain state of a body loaded in more than one direction at once. This article stays inside the "Properties of Areas and Axial Loading" scope of ME10001 — triaxial and biaxial isotropic elastic strain, the EE–GG–ν\nu relation, and allowable stress via factor of safety. Full stress-tensor eigenanalysis and yield criteria are out of scope here.

The core insight

For an isotropic linear-elastic material, the strain along any axis is the direct stress on that axis divided by EE, minus ν/E\nu/E times the sum of the stresses on the other two axes.

The core idea

Stretch a bar along xx with stress σx\sigma_x alone, and it does not just elongate along xx — it contracts sideways. Poisson's ratio quantifies that sideways contraction relative to the axial stretch:

ν=−εlateralεaxial\nu = -\frac{\varepsilon_{lateral}}{\varepsilon_{axial}}

where: εlateral\varepsilon_{lateral} is the strain perpendicular to the applied stress, εaxial\varepsilon_{axial} is the strain along it, and the minus sign encodes that the two strains have opposite sign for a positive ν\nu. Most engineering metals have ν≈0.25\nu \approx 0.25 to 0.350.35; steel is close to ν≈0.3\nu \approx 0.3. Now load a small cube on all three faces at once with normal stresses σx,σy,σz\sigma_x, \sigma_y, \sigma_z (no shear on these faces). Each stress produces a direct strain along its own axis and a Poisson contraction along the other two. Because the material is linear-elastic, these three effects simply add — this superposition is the generalized Hooke's law:

εx=1E[σx−ν(σy+σz)],εy=1E[σy−ν(σz+σx)],εz=1E[σz−ν(σx+σy)]\varepsilon_x = \frac{1}{E}\big[\sigma_x - \nu(\sigma_y+\sigma_z)\big], \qquad \varepsilon_y = \frac{1}{E}\big[\sigma_y - \nu(\sigma_z+\sigma_x)\big], \qquad \varepsilon_z = \frac{1}{E}\big[\sigma_z - \nu(\sigma_x+\sigma_y)\big]

where: EE is Young's modulus, ν\nu is Poisson's ratio, and each strain is built from its own direct term minus the Poisson term pulled from the other two stresses. For biaxial stress (σz=0\sigma_z = 0, the common case of a thin plate loaded in its plane), these reduce to εx=1E(σx−νσy)\varepsilon_x = \frac{1}{E}(\sigma_x-\nu\sigma_y) and εy=1E(σy−νσx)\varepsilon_y = \frac{1}{E}(\sigma_y-\nu\sigma_x) — but εz\varepsilon_z does not vanish: εz=−νE(σx+σy)\varepsilon_z = -\dfrac{\nu}{E}(\sigma_x+\sigma_y). The plate still thins in the thickness direction purely from the Poisson coupling of the in-plane stresses, even though no stress acts there. Shear behaves independently: τxy=Gγxy\tau_{xy} = G\gamma_{xy}, with no coupling to the normal strains for an isotropic material. But EE, GG, and ν\nu are not three independent numbers — for an isotropic material they are locked together by

G=E2(1+ν)G = \frac{E}{2(1+\nu)}

where: GG is the shear modulus. Only two of the three constants need to be known; the third follows from this relation. Design work then reuses the same allowable stress logic from uniaxial loading: each stress component is kept below σallow=σult/FS\sigma_{allow} = \sigma_{ult}/FS, where FSFS is the factor of safety.

Figure
A cube element with normal stresses on three faces and arrows showing direct strain along each axis plus Poisson contraction from the other two
Superposition — each stress stretches its own axis and contracts the other two σx σy σz εx = [σx − ν(σy + σz)] / E, and likewise for εy and εz
Derivation ABuilding the generalized Hooke's law by superposition

The goal is to derive εx=1E[σx−ν(σy+σz)]\varepsilon_x = \frac{1}{E}\big[\sigma_x - \nu(\sigma_y+\sigma_z)\big] for a triaxially loaded isotropic linear-elastic cube starting from nothing more than the uniaxial law σ=Eε\sigma = E\varepsilon and the definition of Poisson's ratio, by applying each stress one at a time and adding the resulting strains along xx.

Step 1. Apply σx\sigma_x alone (imagine σy=σz=0\sigma_y = \sigma_z = 0 for a moment). By the uniaxial law this produces a direct strain along xx of εx(1)=σx/E\varepsilon_x^{(1)} = \sigma_x/E. Because the material contracts sideways under this same stress, it also produces lateral strains along yy and zz of εy(1)=εz(1)=−νεx(1)=−νσx/E\varepsilon_y^{(1)} = \varepsilon_z^{(1)} = -\nu\varepsilon_x^{(1)} = -\nu\sigma_x/E, directly from the definition of ν\nu.

Step 2. Apply σy\sigma_y alone. By the same reasoning it produces a direct strain along yy of σy/E\sigma_y/E, and — this is the term we actually want — a lateral (Poisson) contribution along xx of εx(2)=−νσy/E\varepsilon_x^{(2)} = -\nu\sigma_y/E, since xx is now a "lateral" direction relative to the applied σy\sigma_y.

Step 3. Apply σz\sigma_z alone. Identically, it contributes εx(3)=−νσz/E\varepsilon_x^{(3)} = -\nu\sigma_z/E along xx.

Step 4. Superpose. Because the material is linear-elastic, the true strain along xx under all three stresses acting together is just the sum of the three separate contributions along xx: εx=εx(1)+εx(2)+εx(3)=σxE−νσyE−νσzE=1E[σx−ν(σy+σz)]\varepsilon_x = \varepsilon_x^{(1)} + \varepsilon_x^{(2)} + \varepsilon_x^{(3)} = \dfrac{\sigma_x}{E} - \dfrac{\nu\sigma_y}{E} - \dfrac{\nu\sigma_z}{E} = \dfrac{1}{E}\big[\sigma_x - \nu(\sigma_y+\sigma_z)\big]. Cycling x→y→zx \to y \to z gives the matching expressions for εy\varepsilon_y and εz\varepsilon_z by the same argument with the roles of the axes rotated.

Worked example

A steel plate (E=200E = 200 GPa, ν=0.3\nu = 0.3) carries biaxial in-plane stresses σx=120\sigma_x = 120 MPa and σy=60\sigma_y = 60 MPa (both tension), with σz=0\sigma_z = 0. Find εx\varepsilon_x, εy\varepsilon_y, εz\varepsilon_z, and the plate's shear modulus GG.

Step 1. Compute εx\varepsilon_x using the biaxial form: εx=1E(σx−νσy)=120×106−0.3(60×106)200×109=102×106200×109=5.10×10−4\varepsilon_x = \dfrac{1}{E}(\sigma_x-\nu\sigma_y) = \dfrac{120\times10^6 - 0.3(60\times10^6)}{200\times10^9} = \dfrac{102\times10^6}{200\times10^9} = 5.10\times10^{-4}.

Step 2. Compute εy\varepsilon_y the same way: εy=1E(σy−νσx)=60×106−0.3(120×106)200×109=24×106200×109=1.20×10−4\varepsilon_y = \dfrac{1}{E}(\sigma_y-\nu\sigma_x) = \dfrac{60\times10^6 - 0.3(120\times10^6)}{200\times10^9} = \dfrac{24\times10^6}{200\times10^9} = 1.20\times10^{-4}.

Step 3. Compute εz\varepsilon_z from the full triaxial formula with σz=0\sigma_z = 0: εz=1E[0−ν(σx+σy)]=−0.3(180×106)200×109=−2.70×10−4\varepsilon_z = \dfrac{1}{E}\big[0-\nu(\sigma_x+\sigma_y)\big] = \dfrac{-0.3(180\times10^6)}{200\times10^9} = -2.70\times10^{-4}. Even with no stress in the zz-direction, the plate thins — a direct demonstration of Poisson coupling.

Step 4. Get the shear modulus from G=E/[2(1+ν)]=200×109/[2(1.3)]=7.69×1010G = E/[2(1+\nu)] = 200\times10^9/[2(1.3)] = 7.69\times10^{10} Pa =76.9= 76.9 GPa; only two of EE, GG, ν\nu were independent inputs.

Answer: εx=5.10×10−4\varepsilon_x = 5.10\times10^{-4}, εy=1.20×10−4\varepsilon_y = 1.20\times10^{-4}, εz=−2.70×10−4\varepsilon_z = -2.70\times10^{-4}, G=76.9G = 76.9 GPa.

Competitive stretch

For an aluminium alloy with E=70E = 70 GPa, ν=0.33\nu = 0.33, σult=310\sigma_{ult} = 310 MPa, and a required factor of safety FS=2.5FS = 2.5: the allowable stress is σallow=σult/FS=310/2.5=124\sigma_{allow} = \sigma_{ult}/FS = 310/2.5 = 124 MPa, and treating this as a simple uniaxial design check gives a maximum permissible strain εallow=σallow/E=124×106/70×109=1.77×10−3\varepsilon_{allow} = \sigma_{allow}/E = 124\times10^6/70\times10^9 = 1.77\times10^{-3}, about 0.177%0.177\%.

Common traps

  • Assuming εz=0\varepsilon_z = 0 whenever σz=0\sigma_z = 0 — false in a biaxial stress state; Poisson coupling from the in-plane stresses still produces a strain in the stress-free direction.
  • Forgetting the sign in ν=−εlateral/εaxial\nu = -\varepsilon_{lateral}/\varepsilon_{axial}: a tensile stress in one direction subtracts from strain in the perpendicular directions via the −νσ-\nu\sigma term.
  • Treating EE, GG, ν\nu as three independently measurable constants — for an isotropic material only two are independent, the third follows from G=E/[2(1+ν)]G = E/[2(1+\nu)].
  • Using σult\sigma_{ult} directly as the design limit instead of dividing by the factor of safety to get σallow\sigma_{allow}.

Ask the tutor: if you want to practice going from a given strain state back to the stresses that produced it, or want the sign conventions in the generalized Hooke's law walked through again, ask the AI tutor.

Why does the plate get thinner in the z-direction if there is no stress applied there?

Because Poisson's ratio couples the directions together. The in-plane stresses σx\sigma_x and σy\sigma_y each try to squeeze the material sideways as a side effect of stretching it along their own axis, and zz is a lateral direction relative to both of them. Add up both contractions using the triaxial formula with σz=0\sigma_z = 0 and you get εz=−ν(σx+σy)/E\varepsilon_z = -\nu(\sigma_x+\sigma_y)/E, a small but real negative strain. Nothing pushes on the zz-face directly, but the material still thins there.

If I only know E and nu, can I always find G without a separate test?

Yes, provided the material is isotropic and linear-elastic — that is exactly what G=E/[2(1+ν)]G = E/[2(1+\nu)] gives you. The three constants are not independent for an isotropic material, so measuring any two (usually EE and ν\nu, from a simple tension test) fixes the third. This stops being true for anisotropic materials like fiber composites, where stiffness differs by direction and GG must be measured separately.

Is allowable stress the same idea as before, just applied to each stress component now?

Essentially yes. The factor-of-safety logic from uniaxial loading, σallow=σult/FS\sigma_{allow} = \sigma_{ult}/FS, carries over unchanged: you still divide the material's ultimate strength by a chosen safety margin to get a design limit. In this multiaxial setting you apply that check to each normal stress component individually. Combining several stress components into one equivalent design check (principal or combined stress, yield criteria) is a later-unit refinement, not needed here.

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